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I am overviewing the questions that I am asking and the answers that I am finding in my investigation to interpret the fivefold classification of Sheffer polynomials. I am seeking a combinatorial and algebraic understanding of the classification that would explain how the allowable polynomials manifest conceptual and physical paradigms. Pieces of the Puzzle I am listing out various pieces of the puzzle. Sheffer polynomials Sheffer polynomials are orthogonal polynomial sets {$\{P_n(x)\}$} of the form {$$\sum_{n=0}^{\infty}P_n(x)t^n=A(t)e^{xu(t)}$$} Recurrence relations All orthogonal polynomials satisfy a recurrence relation of the form {$$xP_n(x) = P_{n+1}(x) + A_n P_n(x) + B_n P_{n1}(x)$$} More specifically, Sheffer polynomials satisfy a recurrence relation of the form {$$P_{n+1}(x)=[x(ln+f)]P_n(x)[kn(n1)\gamma n]P_{n1}(x)$$} Here we can assume that {$f$} = 0 because that's equivalent to translating {$x$} to {$xf$}. Solution from recurrence relations As shown by Galiffa and Riston, this recurrence relation is determined by {$a_1$}, {$a_2$}, {$u_2$}, {$u_3$}. Namely, {$f=a_1$}, {$l=2u_2$}, {$k=4u_2^23u_3$}, {$\gamma = a_1^2+2a_22a_1u_2+4u_2^23u_3$} We can turn these equations around to give combinatorial meaning to the initial terms of {$A(t)$} and {$u(t)$}. {$a_1 = f$}, {$a_2 = \frac{f^2+fl+k+\gamma}{2}$}, {$u_2=\frac{l}{2}$}, {$u_3=\frac{l^2k}{3}$}. We can suppose {$f=0$} by considering the translation of {$x$} to {$(xf)$}. Then {$a_2=\frac{k+\gamma}{2}$}. This yields: {$$A(t)=1  \frac{k+\gamma}{2}t^2 + a_3t^3 + a_4^t4 + \cdots$$} {$$u(t)=t  \frac{l}{2}t^2 + \frac{l^2k}{3}t^3 + a_4^t4 + \cdots$$} Note that the higher order terms are not relevant, in that they are determined by the initial terms. Then we get the most natural and profound expressions for {$u'(t)$} and {$A'(t)$}. {$A'(t) = (k+\gamma)t + \textrm{echo terms}$} {$u'(t) = 1  lt + (l^2k)t^2 + \textrm{echo terms}$} Alternatively, factoring {$1+lt+kt^2=(1\alpha t)(1\beta t)$}, we have {$l=(\alpha)+(\beta)$} and {$k=(\alpha)(\beta)$}. Thus we have {$$u'(t) = 1 (\alpha + \beta) t + (\alpha^2 + \alpha\beta + \beta^2)t^2 + \textrm{echo terms}$$} I mean by "echo terms" that the higher order terms are completely determined by the initial terms and have no independent significance. Nand and Nor Note that we can interpret {$l$} to mean {$\textrm{NAND(A,B)}$}, for (NOT (A AND B) = (NOT A) OR (NOT B), and we can interpret {$k$} as {$\textrm{NOR(A,B)}$}, for (NOT (A OR B) = (NOT A) AND (NOT B). NAND and NOR are the two logical operators from which all other operators can be derived. And algebraically, we get an interesting logic where NAND(A,B) = (A)(B) = AB = AND(A,B). I should study this logic. Differential equation {$A(t)$} and {$u(t)$} satisfy the differential equations below, where {$l$}, {$k$}, {$\gamma$} are taken from the recurrence relation above. {$$u'(t)=\frac{1}{1+lt+kt^2}=\frac{A'(t)}{\gamma tA(t)}$$} It is straightforward to solve for {$A(t)$} upon noting that {$\frac{\textrm{d}}{\textrm{dt}}\textrm{ln}A(t)=\frac{A'(t)}{A(t)}$}. I am still working on how to solve for {$u(t)$}. Apparently, it is more sophisticated. I think it depends on inverting {$t(u)$} and noting that {$\frac{\textrm{dt}}{\textrm{du}}=\frac{1}{u'(t)}$}. In the differential equations above, we can factor the quadratic and rewrite them as follows: {$$u'(t)=\frac{1}{(1\alpha t)(1\beta t)}=\frac{1}{1+lt+kt^2}=\frac{A'(t)}{\gamma tA(t)}$$} In the most general case, the solution gives the Meixner polynomials.
Fivefold classification Various specializations of the Meixner polynomials yield a fivefold classification.
Zeng growths Jiang Zeng worked out a combinatorial interpretation of the Sheffer polynomials that is extremely contentful. He and Dongsu Kim write out the recurrence formula much like this: {$$P_{n+1}(x)=[x((u_3+u_4)n+a_1b_1)]P_n(x)n[n+b_11](u_1u_2))P_{n1}(x)$$} It turns out that their expression is overly subtle for our purposes. It is what they need to explain the combinatorics of linearization. Our simpler recurrence relation has everything needed for their combinatorial interpretation of the Sheffer polynomials and even their orthogonality. {$$P_{n+1}(x)=[x(ln+f)]P_n(x)[kn(n1)\gamma n]P_{n1}(x)$$} {$l=u_3+u_4$} and {$k=u_1u_2$} and we know that {$l=\alpha\beta$} and {$k=\alpha\beta$}. Note that the weights {$u_1$}, {$u_2$}, {$u_3$}, {$u_4$} are simply combinatorial place holders. So for our purposes we can set {$u_1=u_3=\alpha$} and {$u_2=u_4=\beta$}. And we suppose {$f=0$}, so we can assume {$a_1=0$}. Walks somewhere and back Zeng's interpretation comes from the recurrence relation. But we can get another recurrence relation by considering the logarithm of the generating function for the Meixner polynomials. Note that {$\textrm{ln}(1\alpha t)=\alpha t + \frac{1}{2}\alpha^2 t^2 + \frac{1}{3}\alpha^3 t^3 + \cdots$} and that {$\beta^{k+1}\alpha^{k+1}=(\beta\alpha)(\beta^k+\beta^{k1}\alpha + \dots + \beta\alpha^{k1} + \alpha^k)$}. {$$\textrm{ln}A(t)+xu(t)=$$} {$$t(x)+\frac{(t)^2}{2}(x(\beta + \alpha)  \gamma)+\frac{(t)^3}{3}(x(\beta^2 + \beta\alpha + \alpha^2)  \gamma(\beta+\alpha)) +$$} {$$\frac{(t)^4}{4}(x(\beta^3 + \beta^2\alpha + \beta\alpha^2 + \alpha^3)  \gamma(\beta^2 + \beta\alpha + \alpha^2))+\dots$$} We can express the walk in {$\alpha$} and {$\beta$} in terms of {$l$} and {$k$}. {$$\sum_{k=0}^{n}\alpha^k\beta^{nk}=\sum_{j=0}^{\frac{n}{2}}(1)^{nj}l^{n2j}k^j\binom{nj}{j}$$} Whereas {$\alpha$} and {$\beta$} were strictly ordered, here we have all possible ways of ordering {$l$} (of length {$1$}) and {$k$} (of length {$2$}). Note that if there are {$j$} segments with weight {$k$} of length {$2$}, then there are {$n2j$} segments with weight {$l$} of length {$1$}, and so there are {$nj$} segments in all, which is why we have {$\binom{nj}{j}$} orderings. Moments {$$L(x^n)=\sum_{\sigma\in S_n}(\alpha)^{a\sigma} (\beta)^{d\sigma} 0^{fix\sigma}(\frac{\gamma}{\alpha\beta})^{cycle\sigma}$$} Schroedinger equation Suspension Given the expression {$$1 = \frac{1\alpha}{1\alpha} = (1\alpha)(1+\alpha+\alpha^2+\alpha^3+\cdots)$$} temporarily substitute {$\alpha=\beta$} in the numerator {$$1 = \frac{1\beta}{1\alpha} = (1\beta)(1+\alpha+\alpha^2+\alpha^3+\cdots)$$} and then consider various degenerate cases such as {$\beta=0$}, {$\beta=\overline{\alpha}$}, {$\beta=\alpha$} and {$\alpha=0$}. Weights
Generating functions
Differential equation
Classical orthogonal polynomials
Lie theory
